How Can Isomorphic Tangent Spaces Help Understand Derivations?

[CMoore]

Hello,

I am trying to understand how the geometric tangent space \mathbb{R}^n_a given by

<br /> \begin{displaymath}<br /> \mathbb{R}^n_a = \{(a,v) | v \in \mathb {R}^n\}<br /> \end{displaymath}<br />

is isomorphic to the space of all derivations of C^{\infty}(\mathbb{R}^n) at a, denoted by T_a(\mathbb{R}^n).

According to the book "Introduction to Smooth Manifolds" by John M. Lee, an isomorphism between these spaces is given by a map that sends each v_a in \mathbb{R}^n_a to the operator that represents the directional derivative evaluated at the point a in the direction of v. If \phi$ denotes the proposed isomorphism, we can write

<br /> \begin{displaymath}<br /> \phi (v_a)(f) = \widetilde{v}_a(f) = v^i \dfrac{\partial f}{\partial x^i}(a) <br /> \end{displaymath}<br />

for any f \in C^{\infty}(\mathbb{R}^n).

To show that the (clearly linear) map \phi is an isomorphism we must show that it is a bijection. Now, to prove that \phi is 1-1, from linear algebra we know that it is sufficient to show that the kernel of \phi contains only the 0 vector in \mathbb{R}^n_a, denoted by 0_a. This means that we must show that the only element of \mathbb{R}^n_a that satisfies \phi (v_a)(f) = 0 is 0_a.
So, suppose
<br /> \begin{displaymath}<br /> v^i \dfrac{\partial f}{\partial x^i}(a) = 0<br /> \end{displaymath}<br />
If we can show that each component v^i = 0 injectivity will follow - and this is where I am stuck. Any ideas?

Thanks

 

[matt grime]

why not choose to evaluate the directional derivatives on nice choices of f, eg the function that picks out components.

 

[CMoore]

Indeed; that is what the author does. He uses the standard coordinate functions for this purpose. But how can such a selection be justified? The relationship should hold not just for particular choices of f but for any f. Obviously, I'm being dense here, but I just can't see it...

 

[matt grime]

Eh? You want to show that the resulting object (thing with \partial_x) is not zero, it suffices to demonstrate so by finding a function, f, on which it doesn't vanish. Since v_i is not zero for some i, it follows that evaluated on the function that gives the i'th coordinate (the derivative is just v_i) is not zero.